66 +48b 3b^2 2)b³+64 b²-4b+16

$$ \frac{66+48b}{b^3+64} \div \frac{3b^2}{b^2-4b+16} = \frac{6(11+8b)}{b^3+4^3} \div \frac{3b^2}{b^2-4b+16} = \frac{6(11+8b)}{(b+4)(b^2-4b+16)} \div \frac{3b^2}{b^2-4b+16} = \frac{6(11+8b)(b^2-4b+16)}{3b^2(b+4)(b^2-4b+16)} = \frac{2(11+8b)}{b^2(b+4)} $$.

Ответ: $$\frac{2(11+8b)}{b^2(b+4)}$$